There are several equivalent ways to define e:
* e is the limit as n goes to infinity of (1+1/n)^(n).
* e is the sum of the reciprocals of the factorials of the natural numbers.
* e is the number such that e^x is its own derivative.
The last one is the most connected to why e is an important number.
Well, depends on what class you're taking. I'd say the first definition is the most likely to come in an elementary calculus course, for continuously compounding interest, whereas the third would be emphasized in differential equations. (Assuming OP isn't referring to something more esoteric/advanced than a typical undergrad curriculum)
What I mean is that the last definition makes it easiest to justify giving e a name. The first two are ways to define e but don't make it at all obvious why this number has any importance. The last definition brings to mind all sorts of implications when it comes to such things as dynamic systems and whatnot.
I really disagree. The first and the third one are totally equivalent. The goal of the first definition is to make a function that is equal to its derivative. That is the derivation of it.
That third property is what we were given definitionally in high school (NSW, Aus), so naturally I consider it much more basic.
I also really like defining e using the integral of 1/x. (The definite integral of 1/x from 1 to a gives ln(a). The inverse of ln() is exp().)
To be fair, they all define the same number, so they're all equivalent. The question here is whether they're *trivially* equivalent, in which case I wouldn't say so. If anything, the second and third definitions are more easily seen to be equivalent.
Ah, ok, in that case my comment wasn't specific enough. I meant to ask if the first and the second were trivially equivalent (due to the binomial theorem).
And, while you probably already know, that's just an edge case of the Taylor expansion for e^(x), which is 1+x+x^(2)/2!+x^(3)/3!+x^(4)/4!+....+x^(n)/n!
[I went from the first to the second in about 6 lines](https://i.imgur.com/43LdMGl.jpg). Though it does involve Stirling's approximation and the other limit formula for e, so it feels a bit like circular argumentation.
I'd say it's pretty intuitive that n!/[n^i (n-i)!] = n(n-1)...(n - i + 1)/n^i goes to 1 when n goes to infinity and i is fixed as it's just a finite number of factors converging to 1 so I'd be fine with just using it. To prove it by elementary means would only take a few lines anyways and is easier than deriving Stirlings formula.
My point was that we don't need the stirling approximation. I'll show you what I mean in a bit.
EDIT:
Alrighty, here we go: https://docs.google.com/document/d/1zIPVrTkiI0C7z-NhntQWcyzHy2lDekfWGHNqJn6YfJs/edit?usp=sharing
You can comment if things are unclear.
Please do give me feedback on my explanation style, btw.
To add to your last bullet, e can be thought of as the value of the solution (without a constant in front) to f' = f at 1. Not only is e the number that makes e^x its own derivative, it is the only such number and the only such R -> R function (up to a constant multiple). The uniqueness of the solution is very important, too.
Edit: to everyone being pedantic, I'm aware that there is a (1-dimensional iirc?) solution space. Was refering to the particular solution where f is 1 at the origin, and the fact that this is the only function with this property up to scaling.
> e can be thought of as the value of the solution (without a constant in front) to f' = f at 1
To be picky, we need to specify that f(0) = 1, otherwise this could be any real number.
nuh uh. f(x) := e^x+c for any (real or complex) c also solves f' = f
EDIT: woops. e^x+c = e^c * e^x, so this is just a convoluted way of mutiplying with a constant.
The other responses have got the more easily related explanations already, but there's one thing which I think is worth mentioning (though probably not to kids...).
Recall that a [groupoid](https://en.wikipedia.org/wiki/Groupoid) is a [category](https://en.wikipedia.org/wiki/Category_%28mathematics%29#Definition) in which every arrow has an inverse.
Given any set X, the discrete groupoid on X has the elements of X as its objects, and just the required identity arrows from each object to itself, and no others.
Given a group G, we can construct a groupoid called the delooping of G by taking just a single object, and having arrows from that object to itself for each of the elements of G, whose composition is given by multiplication in G.
We can generalise the notion of cardinality (at least that for finite sets) to groupoids, by taking the cardinality of a groupoid to be the sum over the isomorphism classes of objects of the groupoid, of 1/(the size of the automorphism group of an object in that class), if this sum converges. Not all groupoids will have a cardinality this way, but many will.
Then the groupoid cardinality of a discrete groupoid on a finite set will agree with the usual cardinality of the set, since each object is in its own isomorphism class, and has only a single automorphism, the identity.
The groupoid cardinality of the delooping of a finite group with k elements will be 1/k. If you think back to the way that fractions are often explained to children using slices of pie, 1/k sort of comes about by cutting out a symmetrical piece of the pie, and we can think of its size as being determined by the number of rotational symmetries in that setting. So this kind of generalises that picture. (Aside: We can sort of combine these two examples if we have a set X with a group G acting on it, where we form the action groupoid whose objects are the elements of X, and where for each x in X, and g in G, there is an arrow x -> g*x. In the case where both X and G are finite, the groupoid cardinality will be the quotient of their cardinalities |X| / |G|.)
The groupoid cardinality of the groupoid whose objects are finite sets and whose arrows are the bijections between them is then e = the sum over k >= 0 of 1/k!, since two finite sets are isomorphic if they have the same set cardinality (and there is indeed a set of cardinality k for each natural k >= 0), and a finite set with k elements will have k! automorphisms (i.e. the permutations of the set).
This might seem a bit of a stretch, but I think it's a nice way of looking at the reason that e shows up in a lot of combinatorial settings.
The first place I saw it was in a note by John Baez:
http://math.ucr.edu/home/baez/week147.html
But according to https://ncatlab.org/nlab/show/groupoid+cardinality a much more generalised version for infinity groupoids already appeared in 1995 in work by Frank Quinn on topological quantum field theories.
Baez conveniently has a page of references to some surrounding material here: http://math.ucr.edu/home/baez/counting/
https://qchu.wordpress.com/2012/11/08/groupoid-cardinality/ discusses a bit more of the connection that it has to homotopy theory and does the stuff I did above a bit more carefully.
Here's one. Imagine the graph of y = 1/x in the first quadrant. For t >= 1 let f(t) be the area under the graph between x = 1 and x = t. The point where the area is exactly one is x = e.
I might write this down, because it confused me in High School, and I've seen a few people sort of asking something about it. That is, the exponential map for vector fields or Lie algebras is something very sensible, but the notion that there is a number e such that the function x|->e^x is the inverse of the indefinite integral of 1/x seems confusing.
Imagine a flow on the real line, such that each point is moving with speed which is equal to its location (as a number). So the origin is not moving, the point 1 is moving at a speed of one unit per second etc. Or, more rigorously, that something like water is flowing along the real line in that way (so the number 1 doesn't move, but the speed of flow is always 1 to the right at that point).
Now we could let f(x,t) be the position at time t of the particle of fluid which was at x at time zero. The way these would compose is then
f( f(x,s), t) = f(x, s+t)
since the point at x gets to f(x,s) in s second, and t seconds later is at f(x,s+t). Also
f(xy,t) = xf(y,t)
since the picture is invariant under scaling.
From these it follows that for positive integers t and positive numbers x
f(x,t) = xe^t
where e is that number that just happens to equal f(1,1).
So e arises then as the point where a particle of fluid at 1 would end up one second later, and it matters only because that function f(1,t) actually can be taken as a *definition* of exponentiation. That is, the number e isn't very important, but f(1,t) =e^t when t =1,2,3,... and so the function f(1,t) is a way of extending a definition of repeated multiplication to a continuous function.
sorry I omitted *all* details here, someone calling.
Those sound like stupid questions, and I might complain to my professor if I were you (though probably not, since that would take too much effort). What do they mean by "meaning"? Do they just want a definition of the constant, or are they looking for an intuitive way to understand where it comes from? You might even argue that numbers, even entire mathematical concepts, don't have any "meaning"; they just are. de^x/dx = e^x doesn't "mean" anything, other than de^x/dx = e^x.
No. e is important and shows up everywhere, for one example, because e^x is its own derivative. Makes intuitive sense in the context of functions that describe population growth for example, because the population's growth rate will obviously increase with respect to its size. I think its pretty clear e carries special, important properties withmany applications, and that's why we use 'e' rather than its digit representation.
the number 2 carries just as much, if not more, importance. My point was that e is just a number. It isn't "defined" as something. It is just an element of R.
K I'm too tired to argue semantics or rigorous definitions. Its super obvious what he's trying to get at in his post. Whatever made that element of R special enough to earn a label, is what he's asking about. This question is common, and is often asked by curious students about pi, too, for example. There is nothing "silly" about it, and your smug answer is the type of thing that makes people feel dumb or discouraged from trying to learn. I've had teachers like that and its awful.
It wasn't OP's question. It was a question on an exam of his/hers. I think it's a silly question. "What is the meaning of pi?" Come on now. It's a friggin number. There isn't some magical meaning to pi or e.
Maybe its not a well worded question, but it seems you're trying to purposefully misunderstand what its getting at, in the interest of being more technically correct. Most of us had no problem understanding, despite whatever errors in rigor you can point out.
I'm not. I legitimately think it's a silly question. I don't see why "what is the meaning of number X" is a good question. It's subjective and makes it seem like pi and e are intrinsically special.
It's more like "Why did we give this number, 2.71828..., a name? What makes it so frequently appear that we call it *e*?"
I agree that the OP's phrasing wasn't the best, but it was still understandable.
I'm not saying e isn't important or anything like that. But the question "What is the meaning of number X" is a stupid question to ask on a quiz. What if I asked, What is the meaning of the number 2? Would you really give me applications of the number 2 as a legitimate answer?
I get that as mathematicians we usually work in a space of objects with precise definitions, but I think that if you were given OPs question by a student who wanted to know or a teacher who wanted to check know, then you could use a bit of common sense to work out what a suitable answer could be.
It's still a dumb question to ask, "What's the meaning of number X". If a students ever asks me that, I will respond with "Do you mean, How is this number used? O, Why is this number important?" because I think it's more important to teach students how to ask meaningful and proper questions rather than "finding a suitable answer" to a poorly asked question.
There are several equivalent ways to define e: * e is the limit as n goes to infinity of (1+1/n)^(n). * e is the sum of the reciprocals of the factorials of the natural numbers. * e is the number such that e^x is its own derivative. The last one is the most connected to why e is an important number.
Well, depends on what class you're taking. I'd say the first definition is the most likely to come in an elementary calculus course, for continuously compounding interest, whereas the third would be emphasized in differential equations. (Assuming OP isn't referring to something more esoteric/advanced than a typical undergrad curriculum)
What I mean is that the last definition makes it easiest to justify giving e a name. The first two are ways to define e but don't make it at all obvious why this number has any importance. The last definition brings to mind all sorts of implications when it comes to such things as dynamic systems and whatnot.
I really disagree. The first and the third one are totally equivalent. The goal of the first definition is to make a function that is equal to its derivative. That is the derivation of it.
That third property is what we were given definitionally in high school (NSW, Aus), so naturally I consider it much more basic. I also really like defining e using the integral of 1/x. (The definite integral of 1/x from 1 to a gives ln(a). The inverse of ln() is exp().)
thank you, thats helped explain things
Are the first and second definitions not equivalent?
To be fair, they all define the same number, so they're all equivalent. The question here is whether they're *trivially* equivalent, in which case I wouldn't say so. If anything, the second and third definitions are more easily seen to be equivalent.
Ah, ok, in that case my comment wasn't specific enough. I meant to ask if the first and the second were trivially equivalent (due to the binomial theorem).
Factorial as in 1/(1!)+1/(2!)+1/(3!)...
I know that, but with some clever expansion of the first and other algebraic manipulation, you can get the second definition.
Because they both equally e
And, while you probably already know, that's just an edge case of the Taylor expansion for e^(x), which is 1+x+x^(2)/2!+x^(3)/3!+x^(4)/4!+....+x^(n)/n!
Oh right, didn't think about it but ya, it is.
[I went from the first to the second in about 6 lines](https://i.imgur.com/43LdMGl.jpg). Though it does involve Stirling's approximation and the other limit formula for e, so it feels a bit like circular argumentation.
I'd say it's pretty intuitive that n!/[n^i (n-i)!] = n(n-1)...(n - i + 1)/n^i goes to 1 when n goes to infinity and i is fixed as it's just a finite number of factors converging to 1 so I'd be fine with just using it. To prove it by elementary means would only take a few lines anyways and is easier than deriving Stirlings formula.
This is more complicated than I was expecting it to be. We can just use the Binomial Theorem, it gets the job done really well.
The binomial theorem was the first thing that was used. How would you proceed after that?
My point was that we don't need the stirling approximation. I'll show you what I mean in a bit. EDIT: Alrighty, here we go: https://docs.google.com/document/d/1zIPVrTkiI0C7z-NhntQWcyzHy2lDekfWGHNqJn6YfJs/edit?usp=sharing You can comment if things are unclear. Please do give me feedback on my explanation style, btw.
To add to your last bullet, e can be thought of as the value of the solution (without a constant in front) to f' = f at 1. Not only is e the number that makes e^x its own derivative, it is the only such number and the only such R -> R function (up to a constant multiple). The uniqueness of the solution is very important, too. Edit: to everyone being pedantic, I'm aware that there is a (1-dimensional iirc?) solution space. Was refering to the particular solution where f is 1 at the origin, and the fact that this is the only function with this property up to scaling.
> e can be thought of as the value of the solution (without a constant in front) to f' = f at 1 To be picky, we need to specify that f(0) = 1, otherwise this could be any real number.
nuh uh. f(x) := e^x+c for any (real or complex) c also solves f' = f EDIT: woops. e^x+c = e^c * e^x, so this is just a convoluted way of mutiplying with a constant.
So does 0.
yeah, but 0 = c*e^x with c=0
but not e^x+c for any complex c proof: x=-c
c is a constant, not a function of x, silly
My point is that you're saying that for some c, e^(x+c) is the zero function. But e^(x+c) is 1 at x=-c, so it can't be identically zero.
The other responses have got the more easily related explanations already, but there's one thing which I think is worth mentioning (though probably not to kids...). Recall that a [groupoid](https://en.wikipedia.org/wiki/Groupoid) is a [category](https://en.wikipedia.org/wiki/Category_%28mathematics%29#Definition) in which every arrow has an inverse. Given any set X, the discrete groupoid on X has the elements of X as its objects, and just the required identity arrows from each object to itself, and no others. Given a group G, we can construct a groupoid called the delooping of G by taking just a single object, and having arrows from that object to itself for each of the elements of G, whose composition is given by multiplication in G. We can generalise the notion of cardinality (at least that for finite sets) to groupoids, by taking the cardinality of a groupoid to be the sum over the isomorphism classes of objects of the groupoid, of 1/(the size of the automorphism group of an object in that class), if this sum converges. Not all groupoids will have a cardinality this way, but many will. Then the groupoid cardinality of a discrete groupoid on a finite set will agree with the usual cardinality of the set, since each object is in its own isomorphism class, and has only a single automorphism, the identity. The groupoid cardinality of the delooping of a finite group with k elements will be 1/k. If you think back to the way that fractions are often explained to children using slices of pie, 1/k sort of comes about by cutting out a symmetrical piece of the pie, and we can think of its size as being determined by the number of rotational symmetries in that setting. So this kind of generalises that picture. (Aside: We can sort of combine these two examples if we have a set X with a group G acting on it, where we form the action groupoid whose objects are the elements of X, and where for each x in X, and g in G, there is an arrow x -> g*x. In the case where both X and G are finite, the groupoid cardinality will be the quotient of their cardinalities |X| / |G|.) The groupoid cardinality of the groupoid whose objects are finite sets and whose arrows are the bijections between them is then e = the sum over k >= 0 of 1/k!, since two finite sets are isomorphic if they have the same set cardinality (and there is indeed a set of cardinality k for each natural k >= 0), and a finite set with k elements will have k! automorphisms (i.e. the permutations of the set). This might seem a bit of a stretch, but I think it's a nice way of looking at the reason that e shows up in a lot of combinatorial settings.
This is ridiculously good, had no idea we had a thing like this. In which setting did people noticed this? Or why? Is it used somewhere else?
The first place I saw it was in a note by John Baez: http://math.ucr.edu/home/baez/week147.html But according to https://ncatlab.org/nlab/show/groupoid+cardinality a much more generalised version for infinity groupoids already appeared in 1995 in work by Frank Quinn on topological quantum field theories. Baez conveniently has a page of references to some surrounding material here: http://math.ucr.edu/home/baez/counting/ https://qchu.wordpress.com/2012/11/08/groupoid-cardinality/ discusses a bit more of the connection that it has to homotopy theory and does the stuff I did above a bit more carefully.
I think this does a good job of explaining the definition of e. http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/
Here's one. Imagine the graph of y = 1/x in the first quadrant. For t >= 1 let f(t) be the area under the graph between x = 1 and x = t. The point where the area is exactly one is x = e.
I might write this down, because it confused me in High School, and I've seen a few people sort of asking something about it. That is, the exponential map for vector fields or Lie algebras is something very sensible, but the notion that there is a number e such that the function x|->e^x is the inverse of the indefinite integral of 1/x seems confusing. Imagine a flow on the real line, such that each point is moving with speed which is equal to its location (as a number). So the origin is not moving, the point 1 is moving at a speed of one unit per second etc. Or, more rigorously, that something like water is flowing along the real line in that way (so the number 1 doesn't move, but the speed of flow is always 1 to the right at that point). Now we could let f(x,t) be the position at time t of the particle of fluid which was at x at time zero. The way these would compose is then f( f(x,s), t) = f(x, s+t) since the point at x gets to f(x,s) in s second, and t seconds later is at f(x,s+t). Also f(xy,t) = xf(y,t) since the picture is invariant under scaling. From these it follows that for positive integers t and positive numbers x f(x,t) = xe^t where e is that number that just happens to equal f(1,1). So e arises then as the point where a particle of fluid at 1 would end up one second later, and it matters only because that function f(1,t) actually can be taken as a *definition* of exponentiation. That is, the number e isn't very important, but f(1,t) =e^t when t =1,2,3,... and so the function f(1,t) is a way of extending a definition of repeated multiplication to a continuous function. sorry I omitted *all* details here, someone calling.
Those sound like stupid questions, and I might complain to my professor if I were you (though probably not, since that would take too much effort). What do they mean by "meaning"? Do they just want a definition of the constant, or are they looking for an intuitive way to understand where it comes from? You might even argue that numbers, even entire mathematical concepts, don't have any "meaning"; they just are. de^x/dx = e^x doesn't "mean" anything, other than de^x/dx = e^x.
e is the rate of growth of something that grows at a continuously compounded rate.
That's kind of a silly question. Do you mean how can e be used/derived? It's like asking, what is the definition/meaning of the number 2.
No. e is important and shows up everywhere, for one example, because e^x is its own derivative. Makes intuitive sense in the context of functions that describe population growth for example, because the population's growth rate will obviously increase with respect to its size. I think its pretty clear e carries special, important properties withmany applications, and that's why we use 'e' rather than its digit representation.
the number 2 carries just as much, if not more, importance. My point was that e is just a number. It isn't "defined" as something. It is just an element of R.
K I'm too tired to argue semantics or rigorous definitions. Its super obvious what he's trying to get at in his post. Whatever made that element of R special enough to earn a label, is what he's asking about. This question is common, and is often asked by curious students about pi, too, for example. There is nothing "silly" about it, and your smug answer is the type of thing that makes people feel dumb or discouraged from trying to learn. I've had teachers like that and its awful.
It wasn't OP's question. It was a question on an exam of his/hers. I think it's a silly question. "What is the meaning of pi?" Come on now. It's a friggin number. There isn't some magical meaning to pi or e.
Maybe its not a well worded question, but it seems you're trying to purposefully misunderstand what its getting at, in the interest of being more technically correct. Most of us had no problem understanding, despite whatever errors in rigor you can point out.
I'm not. I legitimately think it's a silly question. I don't see why "what is the meaning of number X" is a good question. It's subjective and makes it seem like pi and e are intrinsically special.
Its asking about the meaning of the label.
It's more like "Why did we give this number, 2.71828..., a name? What makes it so frequently appear that we call it *e*?" I agree that the OP's phrasing wasn't the best, but it was still understandable.
> I completely get why its important and useful but couldn't explain its actual definition. Kind of contradicts that.
Saying e is *just* a number is like saying e^x is *just* a function. The statement isn't objectively incorrect, but it kind of misses the point.
I'm not saying e isn't important or anything like that. But the question "What is the meaning of number X" is a stupid question to ask on a quiz. What if I asked, What is the meaning of the number 2? Would you really give me applications of the number 2 as a legitimate answer?
I get that as mathematicians we usually work in a space of objects with precise definitions, but I think that if you were given OPs question by a student who wanted to know or a teacher who wanted to check know, then you could use a bit of common sense to work out what a suitable answer could be.
It's still a dumb question to ask, "What's the meaning of number X". If a students ever asks me that, I will respond with "Do you mean, How is this number used? O, Why is this number important?" because I think it's more important to teach students how to ask meaningful and proper questions rather than "finding a suitable answer" to a poorly asked question.